How to show $n \sum_{k>n} (k^2 \log k)^{-1} \sim (\log n)^{-1}$?

Question

How does one show that $$n \sum_{k>n} \frac{1}{k^2 \log k} \sim \frac{1}{\log n} \quad ?$$

Many thanks for your help.

Answer 1

We can expect $\sum_{k>n} \frac{1}{k^2 \log k} \sim \int_n^\infty \frac{1}{x^2\ln x}dx.$ Try L'Hopital on

$$\frac{\int_x^\infty \frac{dt}{t^2\ln t}}{1/(x\ln x)}.$$